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In an L-R-C series circuit, R = 400 Ω, L = 0.350 H, and C = 0.0120 mF. (a) What is the resonance angular frequency of the circuit? (b) The capacitor can withstand a peak voltage of 670 V. If the voltage source operates at the resonance frequency, what maximum voltage amplitude can it have if the maximum capacitor voltage is not exceeded?

Short Answer

Expert verified

(a)The resonance angular frequency of the circuit is and (b) Maximum voltage amplitude it can have if the maximum capacitor voltage is not exceeded is 40.7V.

Step by step solution

01

Step-1: Formulas used  

The angular frequency depends on the inductance and the capacitance and it is given by ω=1LC.

Capacitive reactance is given byXC=1ωC

02

Step-2: Calculation of angular frequency

When the angular frequency is varied, the current amplitude will change and the maximum current is produced at the minimum impedance. The process of peaking the current represents the resonance where the initial angular frequency is the same for the final angular frequency.

Plug the values of L and C into equation

ω=1LC=10.350H0.0120×10-6F=1.54×104rad/s

03

Step-3: (b)Calculation of maximum voltage amplitude

First calculate the capacitive reactance

XC=11.54×104rad/s0.0120×10-6F=5410Ω

The current through the capacitor is given by

I=VcXc=550V5410Ω=0.101A

Now, get the amplitude voltage by

V=0.101A400Ω=40.7V

Hence, (a)The resonance angular frequency of the circuit isω=1.54×104rad/s and (b) Maximum voltage amplitude it can have if the maximum capacitor voltage is not exceeded is 40.7V.

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